C/C++ 移位计算代替乘除运算

测试移位和乘除的比较,发现移位比乘除运算快一个位数的速度,但是难点在于判断是否是2的幂次级的数，如果不是还得通过代码拆分到2的幂次+上分子的累和，然后通过移位得到2的次幂数这样;

下列代码只是简单的判断是幂级数则移位运算，不是则正常计算，不是完全都是移位算，(之后如果找到方法判断更快在更新)

C/C++：

 #include <iostream>

 uint64_t Multiply_Dived(uint64_t member, uint64_t denominator, char style)
 {
     //判断分母是否是2的次幂
     if (denominator & (denominator - ))
     {
         if (style == '*')
         {
             while ((denominator >>= ) != )
             {
                 member <<= ;
             }

         } else
         {
             while ((denominator >>= ) != )
             {
                 member >>= ;
             }

         }
         return member;
     }

     if (style == '*')
     {
         return member * denominator;
     } else
     {
         return member / denominator;
     }
 }

 uint64_t Nomorl(uint64_t member, uint64_t denominator, char style)
 {
     if (style == '*')
     {
         return member * denominator;
     } else
     {
         return member / denominator;
     }
 }

 int main()
 {
     std::clock_t start = , stop = ;
     start = clock();
     for (uint64_t i = ; i < ; i++)
     {
         Multiply_Dived(, , '/');
     }
     stop = clock();
     std::cout << "除移位运行时间: " << ((double) (stop - start) / CLOCKS_PER_SEC) << std::endl;

     start = clock();
     for (uint64_t i = ; i < ; i++)
     {
         Nomorl(, , '/');
     }
     stop = clock();
     std::cout << "正常除运行时间: " << ((double) (stop - start) / CLOCKS_PER_SEC) << std::endl;

     return ;
 }